Discuss the method to determine cell potential of any cell when standard hydrogen electrode is considered as cathode with suitable example.

Explanation

At 298K the emf of the cell, standard hydrogen electrode and second half-cell constructed by taking standard hydrogen electrode as cathode (reference half-cell) and the other half-cell as anode, gives the reduction potential of the other half-cell.
Electrode of the other half-cell as anode || Standard hydrogen electrode as cathode
E.g.,general cell :
`M_((S))|M_((aq))^(n+)(1M)||H_((aq))^(+)(1M)|(1)/(2)|(1)/(2)H_(2(g))(1" bar")|Pt`
and `E_(cell)^(Theta)=(E_(R)^(Theta)-E_(L)^(Theta))` but `E_(R)^(Theta)=0.0V`
`E_(cell)^(Theta)=-E_(L)^(Theta)` (for such cel the cell potential is negative)
e.g., anode of zinc half-cell || cathode of hydrogen electrode half-cell
`Zn_((S))|Zn_((aq))^(2+)(1M)||H_((aq))^(+)(1M)|H_(2(g))(1" bar")|Pt_((S))`
For above cell, practical value of reduction potential is `-0.76V`.
So, for `Zn_((aq))^(2+)(1M)+2e^(-)toZn_((S))`, standard reduction potential `=0.76V`
i.e., `E_(Zn^(2+)|Zn)^(Theta)=-0.76V`
Note: When standard hydrogen electrode is on right side and working as cathode then other half-cell has negative reduction potential value and such cell possess negative emf value.
Reduction reaction occurs on cathode electrode present on right side of cell:
Reaction: `H_("aq, 1M")^(+)+e^(-) to (1)/(2)H_(2(g))(1" bar")`
The negative value of such cell shows oxidation of zinc metal is done by `H^(+)` ions and zinc metal get dissolved in HCl acid. also `H^(+)` ions get reduced by zinc metal.
`Zn_((S))+2H_((aq))^(+) to Zn_((aq))^(2+)+H_(2(g))`